# Homework Help: A problem on triangle and it's perpendicular bisectors.

1. Aug 17, 2013

1. The problem statement, all variables and given/known data

I'm trying to solve the following problem :

In △ABC, coordinates of B are (−3,3). Equation of the perpendicular bisector of side AB is 2x+y−7=0. Equation of the perpendicular bisector of side BC is 3x−y−3=0. Mid point of side AC is E(11/2,7/2). Find AC2.

2. Relevant equations

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3. The attempt at a solution

By solving 3x−y−3=0 and 2x+y−7=0 I find that the intersection of perpendicular bisectors is at (2,3) .
Then using the two points (2,3) and (11/2,7/2), I get the equation of perpendicular bisector of AC as y=x/7+19/7.
So the slope of AC is -7 and then using point slope form , y−7/2=−7(x−11/2) Thus the equation of line AC is y=42−7x .
Similarly equation of line BC is y=2−x/3 .
So AC and BC intersect at (6,0).
By using the fact that E is the midpoint of AC, I find Co-ordinates of A as (5,7).
So the distance between A and C is 5√2, and AC2=50.
But this answer is wrong and the correct answer is 74 ( I checked the answer sheet) .
What have I done wrong ?

2. Aug 17, 2013

### verty

I don't know what went wrong before but here is a slightly different method.

Step 1. Find A using B and the gradient of AB.
Step 2. Find AC^2

I forget about E, E means only A is required.

3. Aug 26, 2013

I don't understand. How can I find AC2 after finding gradient of AB ?

4. Aug 26, 2013

### verty

Your goal is to find A and then AC^2. To find A, find the midpoint of AB then use the midpoint formula.

5. Aug 26, 2013

### haruspex

I agree with your answer, through a slightly different route.
O is the circumcentre, so OC=OB = 5. OE2 = 25/2, so CE2 = 25/2 by Pythagoras.

6. Aug 27, 2013

I tried and got A as (5,7) . Is that right?
Then I get C as (6,0) so AC2=50.

7. Aug 28, 2013

### verty

It must be right, you found the same answer by two different methods and Haruspex found it by a third method. There must have been no mistake originally.